Would unrestricted comprehension and no regularity avoid Russel's paradox with this modification?

Question

Suppose the Set theory in question followed all axioms of ZF or ZFC, except for the axiom of regularity. Additionally the axiom schema of specification was altered from

$$\forall z\forall w_1 \forall w_2 \dots \forall w_n\exists y\forall x[x \in y \iff ((x\in z)\wedge\phi(x))]$$

to unrestricted comprehension

$$\forall w_1 \forall w_2 \dots \forall w_n\exists y\forall x[x \in y \iff ((x\neq y)\wedge\phi(x))]$$

for the purpose of constructing sets beyond just subsets of some $z$.

Without regularity and standard specification, this immediately opens the doors to a possibility of Russel's Paradox. However, a set $R$ cannot be built (this way) to have $R\in R$, as it would have to follow $R \in R \iff ((R\neq R)\wedge\phi(R))$.

Would this avoid Russel's Paradox? Even if so, would it lead to another paradox?

Answer 1

No, this form of comprehension is inconsistent with even the ability to form singletons and unions of two sets.

Suppose $A=\{x: x\not\in x\wedge x\not=A\}$. Then I claim $A\cup\{A\}$ is the usual Russell set.

• If $x\in A\cup \{A\}$ then $x\not\in x$. This is because for such an $x$, either $x\in A$ or $x=A$, and no element of $A$ contains itself by the "$x\not\in x$"-clause of the definition of $A$ and $A\not\in A$ by the "$x\not=A$"-clause of the definition of $A$.

• If $x\not\in x$ then $x\in A\cup\{A\}$. Again, we reason by cases. If $x\not\in x$ then either $x\not=A$ (in which case $x\in A$) or $x=A$ (in which case $x\in\{A\}$), and either way $x\in A\cup\{A\}$.

And now we ask whether $A\cup\{A\}$ is an element of itself.

Answer 2

The above "unrestricted comprehension" schema is equivalent to the statement that there is a single element and that element is empty.

Note that the schema holds in the one element domain where the one element is empty.

There is an empty set. Proof:There is a Y such that x∈Y⟺≠Y∧x≠x. Such a Y must be empty.

Call a set b with exactly 2 elements a 2-set if one element of b is empty and b∉b.

If there are distinct elements, then there is a set which a 2-set or a set whose only element is empty. Poof: Suppose a is empty and a≠b. Then there is a Y such that x∈Y⟺x≠Y∧(x=a∨x=b). Y is not empty and Y∉Y. If b∈Y then Y is a 2-set. If b∉Y then a is the only element of Y.

Let F(x) be the formula ∀t(t∉x∨x∉t). We observe that if F(s) for all s∈x, then F(x).

Suppose that b is not empty. Then there is a set Y whose only element is b.
Proof: There is a Y such that x∈Y⟺x≠Y∧x=b. Y is not empty because if it were then b∈Y. Therefore Y is a set whose only element is b.

Suppose there are distinct elements. Then there is a non-empty set Y such that F(Y).

Proof: If there is a set b whose only element is empty, then F(b). Suppose there is no such set. There is a W such that "x∈W⟺x≠W∧(x is not a 2-set)". W is not empty because there is a 2-set and a set whose only element is a 2-set. Suppose c is empty. There is a set Y such that x∈Y⟺x≠Y∧(x=c∨x=W). Y is a 2-set and F(Y).

There is an A such that x∈A⟺x≠A∧F(x). Then F(A).

There is only one element.
Proof: Suppose there are distinct elements. Then there is a non-empty element b such that F(b). But then there is a Y whose only element is b. One of b and Y must be in A. There is a B such that x∈B⟺x≠B∧(∃t(t∈x)∧((x∈A∨x=A))). A≠B since B only has non-empty elements and so A∈B. But F(B) and therefore B∈A. But this contradicts F(A).